Đặt \(\sqrt[5]{5x+1}=t\Rightarrow x=\frac{t^5-1}{5}\)
\(\lim\limits_{t\rightarrow1}\frac{t-1}{\frac{t^5-1}{5}}=\lim\limits_{t\rightarrow1}\frac{5\left(t-1\right)}{\left(t-1\right)\left(t^4+t^3+t^2+t+1\right)}=\lim\limits_{t\rightarrow1}\frac{5}{t^4+t^3+t^2+t+1}=1\)
Tính giới hạn: \(A=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\sqrt[5]{1-5x}-2017}{x}\)
\(\lim\limits_{x\rightarrow0}\left(\frac{1}{x}+\frac{1}{x^2}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{\left(-x\right)^2+2}}{x-1}\)
\(\lim\limits_{x\rightarrow-\infty}\frac{|x|+\sqrt{x^2+x}}{x+10}\)
\(\lim\limits_{x\rightarrow0}\frac{\sqrt{2x+1}-\sqrt[3]{3x+1}}{x^2}\)
\(\lim\limits_{x\rightarrow0}\frac{\sqrt[2019]{x^3+x^2+x+1}-1}{x}\)
tính \(\lim\limits_{x\rightarrow0}\frac{\sqrt{2x+1}.\sqrt[3]{3x+1}-1}{x}\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-\sqrt[3]{x-1}}{x}\)
b) \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{x-3}+\sqrt[4]{2x-3}}{x-2}\)
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow0^-}\dfrac{2\left|x\right|+x}{x^2-x}\)
b) \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{1+x^4+x^6}}{\sqrt{1+x^3+x^4}}\)
\(\lim\limits_{x\rightarrow1}\frac{x-x^2}{\left(2x-1\right)\left(x^5-3\right)}\)
\(\lim\limits_{x\rightarrow0}x\left(1-\frac{1}{x}\right)\)
Tính
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}.\sqrt{2022x^2+x+1}-1}{x}\)